Linear Transformations Questions and Answers

Answer: C

The dual basis is \(f_{1} = \frac{1}{2}(z_{1} - iz_{2}),\ f_{2} = \frac{1}{2}(-iz_{1} + z_{2})\).

A dual basis must satisfy \(f_{i}(v_{j}) = 1\) if \(i = j\) and 0 otherwise, with \(v_{1} = (1, i)\) and \(v_{2} = (i, 1)\). Checking this pair:

\[ \begin{aligned} f_{1}(1, i) &= \tfrac{1}{2}(1 - i^{2}) = 1 \\ f_{1}(i, 1) &= \tfrac{1}{2}(i - i) = 0 \\ f_{2}(1, i) &= \tfrac{1}{2}(-i + i) = 0 \\ f_{2}(i, 1) &= \tfrac{1}{2}(-i^{2} + 1) = 1 \end{aligned} \]

In the other options at least one condition fails. For instance \(f_{1} = \frac{1}{2}(z_{1} + iz_{2})\) gives \(f_{1}(1, i) = 0\), not 1, ruling out two options; and \(f_{2} = \frac{1}{2}(z_{1} + iz_{2})\) gives \(f_{2}(1, i) = 0\) and \(f_{2}(i, 1) = i\), not 1.

Answer: B

The inverse is x = p + q; y = 2p - q.

Read the given data backwards: \(T^{-1}(1,0) = (1,2)\) and \(T^{-1}(1,1) = (2,1)\). Subtracting, \(T^{-1}(0,1) = (2,1) - (1,2) = (1,-1)\). By linearity

\[ \begin{aligned} T^{-1}(p,q) &= p(1,2) + q(1,-1) \\ &= (p + q,\ 2p - q) \end{aligned} \]

Check with (p, q) = (1, 1): x = 2, y = 1, which is (2, 1) as required. The option x = p - q fails this check, since it gives x = 0, and y = 2p + q gives y = 3, so the other three options are wrong.

Answer: C

T is NEITHER one-one NOR onto.

Find the images of the standard basis by linearity:

\[ \begin{aligned} T(1,0,0) &= (0,1,1) \\ T(0,1,0) &= T(1,1,0)-T(1,0,0) = (1,-1,0) \\ T(0,0,1) &= T(1,1,1)-T(1,1,0) = (0,1,1) \end{aligned} \]

The first and third columns are equal, so T(1,0,-1) = 0 and the kernel is non-trivial: T is not one-one. By rank-nullity, the rank is at most 2 (it is exactly 2, since (0,1,1) and (1,-1,0) are independent), so T is not onto \(\mathbb{R}^{3}\). For a linear map from \(\mathbb{R}^{3}\) to itself, one-one and onto stand or fall together, which rules out the two mixed options.

Answer: A

The characteristic polynomial is \(X^{4} - 8X^{2} + 16\).

Left multiplication by \(A\) acts on each column of \(B\) separately: if \(B = [b_{1}\ b_{2}]\), then \(AB = [Ab_{1}\ Ab_{2}]\). So \(T\) is \(A\) acting on two independent copies of \(\mathbb{C}^{2}\), and its characteristic polynomial is the square of that of \(A\).

\[ \begin{aligned} \det(XI - A) &= X^{2} - 4 \\ \chi_{T}(X) &= (X^{2} - 4)^{2} = X^{4} - 8X^{2} + 16 \end{aligned} \]

\(X^{2} - 4\) is the polynomial of \(A\) itself, but \(T\) acts on a 4-dimensional space so its polynomial has degree 4. \(X^{4} - 16\) has eigenvalues \(\pm 2, \pm 2i\), but \(T\)'s eigenvalues are only \(\pm 2\), each twice.

Answer: D

The FALSE statement is the one about cross products.

Because \(F\) is linear, \(\mathit{vect}(F(P),F(Q)) = F(Q) - F(P) = F(Q - P)\). So the condition says \(\|F(u)\| = \|u\|\) for every vector \(u\): \(F\) is an orthogonal map, \(F^{T}F = I\).

An orthogonal map has trivial kernel, so it is injective, and an injective linear map from \(\mathbb{R}^{3}\) to itself is also surjective. By polarisation, \(u \cdot v = \frac{1}{4}(\|u + v\|^{2} - \|u - v\|^{2})\), so preserving lengths preserves dot products. Those three statements are true.

Cross products are not preserved. In general \(F(u) \times F(v) = (\det F)\,F(u \times v)\), which is not \(u \times v\). Take \(F\) the rotation by \(\frac{\pi}{2}\) about the \(z\)-axis, so \(F(e_2) = -e_1\) and \(F(e_3) = e_3\). Then \(F(e_2) \times F(e_3) = -e_1 \times e_3 = e_2\), but \(e_2 \times e_3 = e_1\).

Answer: C

The matrix is a rotation by angle \(\pi/4\) around the y-axis.

The second basis vector \(e_{2} = (0, 1, 0)\) is left fixed, so the axis is the y-axis. In the xz-plane the matrix acts as \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\) with \(\cos\theta = \sin\theta = 1/\sqrt{2}\), which is \(\theta = \pi/4\). Its determinant is \(\frac{1}{2} + \frac{1}{2} = 1\) and its columns are orthonormal, so it is a proper rotation.

The direction (clockwise or anticlockwise) depends on the orientation convention for the axis, but the angle is \(\pi/4\) in either reading. A rotation about the x-axis would fix \(e_{1}\), and one about the z-axis would fix \(e_{3}\), neither of which happens here; \(\pi/2\) would need \(\cos\theta = 0\).

Answer: B

The matrix is \(\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\).

Apply T to each basis vector and write the result in the basis \(u_{1} = (1, 1)\), \(u_{2} = (1, -1)\). The coefficients form the columns.

\[ \begin{aligned} T(1, 1) &= (-1, 1) = 0 \cdot u_{1} - 1 \cdot u_{2} \\ T(1, -1) &= (1, 1) = 1 \cdot u_{1} + 0 \cdot u_{2} \end{aligned} \]

So the first column is \((0, -1)^{t}\) and the second is \((1, 0)^{t}\).

\(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) is the matrix of T in the standard basis; the change of basis reverses the sense because \(u_{1}, u_{2}\) has the opposite orientation. \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) and diag(1, -1) are reflections, with determinant \(-1\), whereas T is a rotation with determinant 1 in every basis.

Answer: B

The matrix is the one with rows [0 1 0 0 0], [0 0 1 0 0], [0 0 0 1 0], [0 0 0 0 1], [0 0 0 0 0], ones just above the diagonal.

Apply \(L(f) = (f(t) - f(0))/t\) to each basis vector: \(L(1) = 0\), \(L(t) = 1\), \(L(t^{2}) = t\), \(L(t^{3}) = t^{2}\), \(L(t^{4}) = t^{3}\). The coordinates of each image form a column of the matrix: column 1 is zero, column 2 is \(e_{1}\), column 3 is \(e_{2}\), column 4 is \(e_{3}\) and column 5 is \(e_{4}\). Writing these columns side by side gives the superdiagonal of ones.

The first option is its transpose (it would send \(t^{k}\) to \(t^{k+1}\), a shift up in degree), and the other two permute basis vectors, which \(L\) does not do.

Answer: D

T is nilpotent.

T is the backward shift: \(e_{n} \mapsto e_{n-1} \mapsto \cdots \mapsto e_{1} \mapsto 0\). Applying it \(n\) times sends every basis vector to 0, so \(T^{n} = 0\).

It is linear by construction, since it is defined on a basis and extended linearly. It is not idempotent: \(T^{2}e_{3} = e_{1}\) while \(Te_{3} = e_{2}\) (for \(n \ge 3\); for \(n = 2\), \(T^{2} = 0 \ne T\)). Its kernel contains \(e_{1}\), so Ker T is not {0}.

Answer: B

The matrix is \(\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \end{pmatrix}\).

A map from \(\mathbf{R}^{3}\) to \(\mathbf{R}^{2}\) has a \(2 \times 3\) matrix whose columns are the images of the standard basis vectors:

\[ \begin{aligned} T(e_{1}) &= (1, 0) \\ T(e_{2}) &= (0, 1) \\ T(e_{3}) &= (1, -1) \end{aligned} \]

Reading the rows off \(x + z\) and \(y - z\) gives the same matrix. The \(3 \times 2\) options are transposes (they would map \(\mathbf{R}^{2}\) to \(\mathbf{R}^{3}\)), and the other \(2 \times 3\) option swaps the roles of \(x\) and \(y\).